6 1.2 1.346395 ?y'??y?x?15、已知?，0?x?1,取h?0.1，用Euler公式求各点上的近似值
y?1?0解：取步长h?0.1x0?0,xj?x0?jh?0.1j(j?0,1,?,10)
f(x,y)??y?x?1,由Euler公式得
故f(xj,yj)??yj?xj?1
?y0?1??yj?1?yi?hf(xj,yj)??y0?1 即:??yj?1?yj?0.1(?yj?xj?1)?y0?1 ?y?0.9y?0.1x?0.1jj?j?1计算结果列于下表
xj 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 yj 1.00000 1.00000 1.01000 1.02900 1.0561 1.09049 1.13144 1.17830 1.23047 xj 0.9 1.0 yj 1.28742 1.34868 6、取步长h=0.1，用Euler法求解如下初值问题并与精确解y?1?2x进行比较.
2x?dy?y?0?x?1? y?dx?y(0)?1?解：取h?0.1,计算x?[0,1]上结果，此时 Euler法：yn?1?yn?0.1(yn?2xn)yn(n?0,1,2,...)?1?yn?1?yn?(k1?k2)2??2xn改进的Euler法：)?k1?0.1(yn?yn??2(xn?0.1))?k2?0.1(yn?k1?y?kn1?计算结果如下表所示：
31
(n?0,1,2,...) x Euler法y 改进的Euler法y 精确解 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
1.000000 1.000000 1.191818 1.277438 1.358213 1.435133 1.508966 1.580338 1.649783 1.717779 1.784770 1.000000 1.095909 1.184097 1.266201 1.343360 1.416402 1.485956 1.552514 1.616475 1.678166 1.737867 1.000000 1.095445 1.183216 1.264911 1.341641 1.414214 1.483240 1.549193 1.612452 1.673320 1.732051 32